<!DOCTYPE html>
<html lang="en-US">
<!--********************************************-->
<!--*       Generated from PreTeXt source      *-->
<!--*                                          *-->
<!--*         https://pretextbook.org          *-->
<!--*                                          *-->
<!--********************************************-->
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta name="robots" content="noindex, nofollow">
</head>
<body class="ignore-math">
<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution:</dfn> <span class="process-math">\(x=0\)</span> is an ordinary point. According to the theorem, the general solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
y=\sum_{n=0}^{\infty} a_n x^n=a_0 y_1(x)+a_1 y_2(x).\tag{5.3.3}
\end{equation}
</div>
<p class="continuation">Substituting the solution into the ODE:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;(1-x^2) \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2}-2 x \sum_{n=0}^{\infty} n a_n x^{n-1}+\alpha (\alpha+1) \sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2}-\sum_{n=0}^{\infty} n(n-1) a_n x^n-2 \sum_{n=0}^{\infty} n a_n x^n+\alpha (\alpha+1) \sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{k=-2}^{\infty} (k+2)(k+1) a_{k+2} x^k+\sum_{n=0}^{\infty} [-n(n-1)-2 n+\alpha (\alpha+1)] a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+\sum_{n=0}^{\infty} [-n(n-1)-2 n+\alpha (\alpha+1)] a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} \left[(n+2)(n+1) a_{n+2}+\left(-n(n-1)-2 n+\alpha (\alpha+1)\right) a_n \right] x^n=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which gives the recurrence relation as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
a_{n+2}=\frac{-n(n+1)+\alpha (\alpha+1)}{-(n+2)(n+1)} a_n=-\frac{(\alpha-n)(\alpha+n+1)}{(n+2)(n+1)}a_n.\tag{5.3.4}
\end{equation}
</div>
<span class="incontext"><a href="sec5_3.html#p-221" class="internal">in-context</a></span>
</body>
</html>
